The scatter plot shows the relationship between the test scores of a group of students and the number of hours they spend on social media in a week:

part a: what is the group of points labeled a called? what is the point labeled b called? give a possible reason for the presence of point b. (3 points)
part b: describe the association between a student’s test scores and the number of hours spent on social media. (2 points)

The other answer is amazing

Step-by-step explanation:

im just answering so you can give them brainlist

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.